What is the integral of (6x+7)/(x^2+4x+29) ?

The integral of (6x+7)/(x^2+4x+29) is 3ln| 1/25 (x+2)^2+1|-arctan(1/5 (x+2))+C

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integral of (6x+7)/(x^2+4x+29)

\int \frac{6 x + 7}{x ^{2} + 4 x + 29} d x

3 ln \frac{1}{25} (x + 2)^{2} + 1 - arctan (\frac{1}{5} (x + 2)) + C

Solution steps

\int \frac{6 x + 7}{x ^{2} + 4 x + 29} d x

Complete the square

x^{2} + 4 x + 29 : (x + 2)^{2} + 25

= \int \frac{6 x + 7}{( x + 2 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{6 u - 5}{u ^{2} + 25} d u

= \int \frac{6 u - 5}{u ^{2} + 25} d u

Apply Integral Substitution

= \int \frac{6 v - 1}{v ^{2} + 1} d v

Expand

\frac{6 v - 1}{v ^{2} + 1} : \frac{6 v}{v ^{2} + 1} - \frac{1}{v ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{6 v}{v ^{2} + 1} d v - \int \frac{1}{v ^{2} + 1} d v

\int \frac{6 v}{v ^{2} + 1} d v = 3 ln v^{2} + 1

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 3 ln v^{2} + 1 - arctan (v)

Substitute back

= 3 ln (\frac{x + 2}{5})^{2} + 1 - arctan (\frac{x + 2}{5})

Simplify

3 ln (\frac{x + 2}{5})^{2} + 1 - arctan (\frac{x + 2}{5}) : 3 ln \frac{1}{25} (x + 2)^{2} + 1 - arctan (\frac{1}{5} (x + 2))

= 3 ln \frac{1}{25} (x + 2)^{2} + 1 - arctan (\frac{1}{5} (x + 2))

Add a constant to the solution

= 3 ln \frac{1}{25} (x + 2)^{2} + 1 - arctan (\frac{1}{5} (x + 2)) + C

\frac{d}{d x} (\frac{sin ^{2} ( x )}{cos ( x )})

(- x + 2), (- x^{2} - 2 x), - 2, - 1

(1 + x)^{- \frac{2}{3}}

y = (1 + 2 x)^{10}, (0, 1)

5 x arcsin (x)

What is the integral of (6x+7)/(x^2+4x+29) ?
The integral of (6x+7)/(x^2+4x+29) is 3ln| 1/25 (x+2)^2+1|-arctan(1/5 (x+2))+C